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An $2\sqrt{k}$-approximation algorithm for minimum power $k$ edge disjoint $st$ -paths

Zeev Nutov

TL;DR

This work gives a $2\sqrt{2k}$-approximation algorithm for general costs and it was an open question whether it admits approximation ratio sublinear in $k$ even for unit costs.

Abstract

In minimum power network design problems we are given an undirected graph $G=(V,E)$ with edge costs $\{c_e:e \in E\}$. The goal is to find an edge set $F\subseteq E$ that satisfies a prescribed property of minimum power $p_c(F)=\sum_{v \in V} \max \{c_e: e \in F \mbox{ is incident to } v\}$. In the Min-Power $k$ Edge Disjoint $st$-Paths problem $F$ should contains $k$ edge disjoint $st$-paths. The problem admits a $k$-approximation algorithm, and it was an open question whether it admits approximation ratio sublinear in $k$ even for unit costs. We give a $2\sqrt{2k}$-approximation algorithm for general costs.

An $2\sqrt{k}$-approximation algorithm for minimum power $k$ edge disjoint $st$ -paths

TL;DR

This work gives a -approximation algorithm for general costs and it was an open question whether it admits approximation ratio sublinear in even for unit costs.

Abstract

In minimum power network design problems we are given an undirected graph with edge costs . The goal is to find an edge set that satisfies a prescribed property of minimum power . In the Min-Power Edge Disjoint -Paths problem should contains edge disjoint -paths. The problem admits a -approximation algorithm, and it was an open question whether it admits approximation ratio sublinear in even for unit costs. We give a -approximation algorithm for general costs.
Paper Structure (3 sections, 6 theorems, 4 equations, 1 figure)

This paper contains 3 sections, 6 theorems, 4 equations, 1 figure.

Key Result

Theorem 1

Min-Power $k$-EDP admits a $2\sqrt{2k}$-approximation algorithm.

Figures (1)

  • Figure 1: $4$ disjoint $st$-paths in $F_4$ and the graph $G_3$.

Theorems & Definitions (6)

  • Theorem 1
  • Theorem 2
  • Corollary 3
  • Lemma 4
  • Lemma 5
  • Lemma 6